Question 4.T.8: (Root Test) Given a series ∑xn, let r = lim sup ^n√|xn|. The...

(Root Test)

Given a series xn\sum{x_{n}} , let

r=limsupxnn.r = \lim \sup \sqrt[n]{|xn|}.

Then

(i) xn\sum{x_{n}}  is absolutely convergent if r < 1.

(ii) xn\sum{x_{n}}  is divergent if r > 1.

(iii) The test is inconclusive if r = 1.

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(i) If r < 1, choose a positive number c ∈ (r, 1). Let ε = c − r. By Theorem 3.16(i), there is a positive integer N such that

n ≥ N ⇒ xnn<r+ε=c\sqrt[n]{|xn|} < r + ε = c

xn<cn.|xn| < c^{n}.

Since the geometric series cn\sum{c^{n}} converges, xn\sum|{x_{n}}| by comparison, and therefore xn\sum{x_{n}} is absolutely convergent.

(ii) By Theorem 3.16(iv), there is a subsequence (xnk)(x_{n_{k}}) of (xn)(x_{n}) such that

xnknkr.\sqrt[n_{k}]{|x_{n_{k}}|} → r.

Since r > 1 there is a positive integer N such that

k ≥ N ⇒ xnknk >1\sqrt[n_{k}]{|x_{n_{k}}|}  > 1

xnk>1,|x_{n_{k}}| > 1,

and we conclude that the condition xn0,x_{n} → 0, which is a necessary condition for convergence, is not satisfied.

(iii) In Example 4.3 we found that 1/n\sum{1/n} was divergent, and in Example 4.7 we established the convergence of 1/n2.\sum{1/n^{2}}. But in both cases r=lim1/npn=1.r = \lim \sqrt[n]{1/n^{p}} = 1.

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